Post on 18-Sep-2020
Electronic Instrumentation Experiment 2
* Part A: Intro to Transfer Functions and AC Sweeps * Part B: Phasors, Transfer Functions and Filters * Part C: Using Transfer Functions and RLC Circuits * Part D: Equivalent Impedance and DC Sweeps
Part A Introduction to Transfer
Functions and Phasors Complex Polar Coordinates Complex Impedance (Z) AC Sweeps
Transfer Functions
The transfer function describes the behavior of a circuit at Vout for all possible Vin.
in
out
VVH ≡
Simple Example 321
32*RRR
RRVV inout +++
=
kkkkkVV inout 321
32*++
+=
65
=≡in
out
VVH
VktVtVthen
VktVtVif
out
in
10)2
2sin(5)(
12)2
2sin(6)(
++=
++=
π
π
More Complicated Example
What is H now?
H now depends upon the input frequency (ω = 2πf) because the capacitor and inductor make the voltages change with the change in current.
How do we model H? We want a way to combine the effect of
the components in terms of their influence on the amplitude and the phase.
We can only do this because the signals are sinusoids • cycle in time • derivatives and integrals are just phase
shifts and amplitude changes
We will define Phasors
A phasor is a function of the amplitude and phase of a sinusoidal signal
Phasors allow us to manipulate sinusoids in terms of amplitude and phase changes.
Phasors are based on complex polar coordinates.
Using phasors and complex numbers we will be able to find transfer functions for circuits.
),( φAfV =
Review of Polar Coordinates
point P is at ( rpcosθp , rpsinθp )
22
1tan
PPP
P
PP
yxr
xy
+=
= −θ
Review of Complex Numbers
zp is a single number represented by two numbers zp has a “real” part (xp) and an “imaginary” part (yp)
jj
jjj
−=
−=⋅−≡
11
1
Complex Polar Coordinates
z = x+jy where x is A cosφ and y is A sinφ ωt cycles once around the origin once for each
cycle of the sinusoidal wave (ω=2πf)
Now we can define Phasors
.),(sincos,
)sin()cos(,)cos()(
droppedisitsotermeachtocommonistjAAVsimplyor
tjAtAVletthentAtVif
ωφφφωφω
φω
+=
+++=
+=
The real part is our signal. The two parts allow us to determine the
influence of the phase and amplitude changes mathematically.
After we manipulate the numbers, we discard the imaginary part.
The “V=IR” of Phasors
The influence of each component is given by Z, its complex impedance
Once we have Z, we can use phasors to analyze circuits in much the same way that we analyze resistive circuits – except we will be using the complex polar representation.
ZIV
=
Magnitude and Phase
VofphasexyV
VofmagnitudeAyxV
jyxAjAV
φ
φφ
=
=∠
=+≡
+=+≡
−1
22
tan
sincos
Phasors have a magnitude and a phase derived from polar coordinates rules.
Influence of Resistor on Circuit
Resistor modifies the amplitude of the signal by R
Resistor has no effect on the phase
RIV RR =
)sin(*)()sin()(
tARtVthentAtIif
R
R
ωω
==
Influence of Inductor on Circuit
Inductor modifies the amplitude of the signal by ωL
Inductor shifts the phase by +π/2
dtdILV L
L =
)2
sin(*)(
)cos(*)()sin()(
πωω
ωωω
+=
==
tALtVor
tALtVthentAtIif
L
L
L
Note: cosθ=sin(θ+π/2)
Influence of Capacitor on Circuit
Capacitor modifies the amplitude of the signal by 1/ωC
Capacitor shifts the phase by -π/2
∫= dtIC
V CC1
)2
sin(*1)2
sin(*1)(
)cos(*1)cos(*1)(
)sin()(
πωω
ππωω
πωω
ωω
ω
−=−+=
−=−
=
=
tAC
tAC
tVor
tAC
tAC
tVthen
tAtIif
C
C
C
Understanding the influence of Phase
=∠ −
xyV 1tan
180
)(0tan
00:
9020
tan
00:
9020
tan
00:
00tan
00:
1
1
1
1
±=
−=
=∠
<=−
−=−=
=∠
<=−
==
=∠
>=+
=
=∠
>=+
−−
−−
+−
+−
ππ
π
π
orx
Vthen
xandyifreal
yVthen
yandxifj
yVthen
yandxifjx
Vthen
xandyifreal
Complex Impedance
Z defines the influence of a component on the amplitude and phase of a circuit • Resistors: ZR = R
• change the amplitude by R
• Capacitors: ZC=1/jωC • change the amplitude by 1/ωC • shift the phase -90 (1/j=-j)
• Inductors: ZL=jωL • change the amplitude by ωL • shift the phase +90 (j)
ZIV
=
AC Sweeps AC Source sweeps from 1Hz to 10K Hz
Transient at 10 Hz Transient at 100 Hz Transient at 1k Hz
Notes on Logarithmic Scales
Capture/PSpice Notes Showing the real and imaginary part of the signal
• in Capture: PSpice->Markers->Advanced • ->Real Part of Voltage • ->Imaginary Part of Voltage
• in PSpice: Add Trace • real part: R( ) • imaginary part: IMG( )
Showing the phase of the signal • in Capture:
• PSpice->Markers->Advanced->Phase of Voltage
• in PSPice: Add Trace • phase: P( )
Part B Phasors Complex Transfer Functions Filters
Definition of a Phasor
φφφω
sincos,)cos()(
jAAVletthentAtVif
+=
+=
The real part is our signal. The two parts allow us to determine the
influence of the phase and amplitude changes mathematically.
After we manipulate the numbers, we discard the imaginary part.
Phasor References
http://ccrma-www.stanford.edu/~jos/filters/Phasor_Notation.html
http://www.ligo.caltech.edu/~vsanni/ph3/ExpACCircuits/ACCircuits.pdf
http://ptolemy.eecs.berkeley.edu/eecs20/berkeley/phasors/demo/phasors.html
Magnitude and Phase
VofphasexyV
VofmagnitudeAyxV
jyxAjAV
φ
φφ
=
=∠
=+≡
+=+≡
−1
22
tan
sincos
Phasors have a magnitude and a phase derived from polar coordinates rules.
Euler’s Formula
θθθ sincos je j +=
2132
13
)(
2
1
2
1
2
13
,
sincos
21
2
1
θθθ
θθ
θθθ
θ
θ
−==
===
=+=+=
−
andrrrtherefore
err
erer
zzzthen
rejrrjyxzif
jj
j
j
214214
)(2121214
,
2121
θθθ
θθθθ
+=⋅=⋅=⋅=⋅= +
andrrrthereforeerrererzzzand jjj
Manipulating Phasors (1)
2132
13
)(
2
1
2
1)(
2
)(1
2
13
)(
,
)sin()cos(
21
2
1
2
1
φφ
φωφω
φφφ
φ
ω
ω
φω
φω
φω
−=∠=
====
=+++=
−+
+
+
XandAAXtherefore
eAA
ee
ee
AA
eAeA
VVX
AetjtAV
jj
tj
tj
tj
tj
tj
Note ωt is eliminated by the ratio • This gives the phase change between
signal 1 and signal 2
Manipulating Phasors (2)
22
22
21
2
2
13
1
yx
yx
V
VX
+
+==
333
222111
jyxVjyxVjyxV
+=
+=+=
−
=∠−∠=∠ −−
2
21
1
11213 tantan
xy
xyVVX
Complex Transfer Functions
If we use phasors, we can define H for all circuits in this way.
If we use complex impedances, we can combine all components the way we combine resistors.
H and V are now functions of j and ω
)()()(
ωωω
jVjVjH
in
out
≡
Complex Impedance
Z defines the influence of a component on the amplitude and phase of a circuit • Resistors: ZR = R • Capacitors: ZC=1/jωC • Inductors: ZL=jωL
We can use the rules for resistors to analyze circuits with capacitors and inductors if we use phasors and complex impedance.
ZIV
=
Simple Example
( ) ( )
11)(1
1
)(
)()()(
+=⋅
+=
+=
+==
RCjjH
CjCj
CjR
CjjH
ZZZ
IZZIZ
jVjVjH
CR
C
CR
C
in
out
ωω
ωω
ω
ωω
ωωω
CjZ
RZ
C
R
ω1
=
=
Simple Example (continued)
11)(
+=
RCjjH
ωω
222
22
)(11
)(101
101
)(RCRCRCj
jjH
ωωωω
+=
+
+=
++
=
)(tan1
tan10tan)(
)1()01()(
111 RCRCjH
RCjjjH
ωωω
ωω
−−− −=
−
=∠
+∠−+∠=∠
2)(11)(
RCjH
ωω
+= )(tan)( 1 RCjH ωω −−=∠
High and Low Pass Filters High Pass Filter
H = 0 at ω → 0
H = 1 at ω → ∞
Η = 0.707 at ωc
Low Pass Filter
H = 1 at ω → 0
H = 0 at ω → ∞
Η = 0.707 at ωc
ωc=2πfc
fc
fc
ωc=2πfc
Corner Frequency The corner frequency of an RC or RL circuit
tells us where it transitions from low to high or visa versa.
We define it as the place where
For RC circuits:
For RL circuits:
21)( =cjH ω
LR
c =ω
RCc1
=ω
Corner Frequency of our example
( )212 RCω+= ( )2
21 ω=
RC
RCjjH
ωω
+=
11)(
21)( =ωjH
21
)(11
21
)(11)( 22
=+
=+
=RCRC
jHωω
ω
RCc1
=ω
H(jω), ωc, and filters We can use the transfer function, H(jω), and the
corner frequency, ωc, to easily determine the characteristics of a filter.
If we consider the behavior of the transfer function as ω approaches 0 and infinity and look for when H nears 0 and 1, we can identify high and low pass filters.
The corner frequency gives us the point where the filter changes: π
ω2
ccf =
Taking limits
012
2
012
2)(bbbaaajH
++++
=ωωωωω
At low frequencies, (ie. ω=10-3), lowest power of ω dominates
At high frequencies (ie. ω =10+3), highest power of ω dominates
0
00
03
16
2
00
31
62
101010101010)(
ba
bbbaaajH ≈
++++
= −−
−−
ω
2
20
03
16
2
00
31
62
101010101010)(
ba
bbbaaajH ≈
++++
= ++
++
ω
Taking limits -- Example
523159)( 2
2
+++
=ωω
ωωωjH
At low frequencies, (lowest power)
At high frequencies, (highest power)
ωωω 35
15)( ==jH LO
339)( 2
2
==ωωωjH HI
Our example at low frequencies
)(010tan)(
110)(
1 axisxonjH
asjH
LOW
LOW
+=
=∠
==→
−ω
ωω
RCjjH
ωω
+=
11)(
101
1)( =+
=ωjH LOW
Our example at high frequencies
220
0tan
10tan)(
011)(
11 ππωω
ωωω
−=−=
−
=∠
=∞
==∞→
−− RCjH
RCjasjH
HIGH
HIGH
RCjjH
ωω
+=
11)(
RCjjH HIGH ωω 1)( =
Our example is a low pass filter
RCf c
c ππω
21
2==
01 == HIGHLOW HH
What about the phase?
Our example has a phase shift
0
1
=
=
HIGH
LOW
H
H
90)(
0)(
−=∠
=∠
ω
ω
jHjH
HIGH
LOW
Part C Using Transfer Functions Capacitor Impedance Proof More Filters Transfer Functions of RLC Circuits
Using H to find Vout
in
out
in
out
jin
jout
tjjin
tjjout
in
out
eAeA
eeAeeA
VVjH
φ
φ
ωφ
ωφ
ω ===
)(
inout
inout
jin
jHjjout
jin
jout
eAejHeA
eAjHeAφωφ
φφ
ω
ω)()(
)(∠=
=
inout AjHA )( ω= inout jH φωφ +∠= )(
Simple Example (with numbers)
157.0)2(1
1)(2
2
=+
=π
ωjH 41.11
2tan0)( 1 −=
−=∠ − πωjH
)4
2cos(2)(
11ππ
µ
+=
Ω==
tkVtV
kRFC
in
)41.1785.02cos(2*157.0)( −+= tkVtVout π
121
11121
11)(
+=
+=
+=
jkkjRCjjH
πµπωω
)625.02cos(314.0)( −= tkVtVout π
Capacitor Impedance Proof
)(1)()()()(
)()cos()(Re)(
)()(
)sin()cos()(
)cos()()()(
)()(
)(
tICj
tVtVCjdt
tdVCtI
tVjtAjdt
jVddt
tdV
jVjeAjdt
dAedt
jVdAetjAtAjV
tAtVanddt
tdVCtI
CCCC
C
CCC
Ctj
tjC
tjC
CC
C
ωω
ωφωωω
ωωωω
φωφωω
φω
φωφω
φω
===
=+=
=
===
=+++=
+==
++
+
CjZC ω
1=Prove:
Band Filters
f0
f0
Band Pass Filter
H = 0 at ω → 0
H = 0 at ω → ∞
Η = 1 at ω0=2πf0
Band Reject Filter
H = 1 at ω → 0
H = 1 at ω → ∞
Η = 0 at ω0 =2πf0
Resonant Frequency The resonant frequency of an RLC circuit tells
us where it reaches a maximum or minimum. This can define the center of the band (on a band
filter) or the location of the transition (on a high or low pass filter).
The equation for the resonant frequency of an RLC circuit is:
LC1
0 =ω
Another Example
11
1
1
)( 22 ++=
++=
LCjRCjCj
LjR
CjjHωω
ωω
ωω
CjZ
LjZRZ
C
LR
ω
ω1
=
==
RCjLCjH
ωωω
+−=
)1(1)( 2
At Very Low Frequencies
0)(10)(
111)(
=∠
=→
==
ω
ωω
ω
jHjH
jH
LOW
LOW
LOW
At Very High Frequencies
ππω
ωω
ωω
−=∠
=∞
=∞→
−=
orjH
jH
LCjH
HIGH
HIGH
HIGH
)(
01)(
1)( 2
At the Resonant Frequency
LC1
0 =ωRCjLC
jHωω
ω+−
=)1(1)( 2
+−
=
+
−
=
LCRCjRC
LCjLC
LC
jH)11(
11)11(
1)( 20ω
2)(
)(
)(
0
0
0
πω
ω
ω
−=∠
=
−=
jH
RCLCjH
RCLCjjH
if L=1mH, C=0.1uF and R=100Ω
ω0=100k rad/sec f0=16k Hz
|Η0|=1
LCf
π21
0 =
radiansH2π
−=∠
Our example is a low pass filter Phase
φ = 0 at ω → 0
φ = -180 at ω → ∞
Magnitude
Η = 1 at ω → 0
Η = 0 at ω → ∞
f0=16k Hz
1
−90
Actual circuit resonance is only at the theoretical resonant frequency, f0, when there is no resistance.
Part D
Equivalent Impedance Transfer Functions of More
Complex Circuits
Equivalent Impedance
Even though this filter has parallel components, we can still handle it.
We can combine complex impedances like resistors to find the equivalent impedance of the components combined.
Equivalent Impedance
LCLj
LCjLj
CjLj
CjLj
ZZZZZ
CL
CLCL 222 111
1
ωω
ωω
ωω
ωω
−=
+=
+
⋅=
+⋅
=
Determine H
LCLjZCL 21 ω
ω−
=
LjLCRLjjH
ωωωω
+−=
)1()( 2
LCLCbymultiply
LCLjR
LCLj
jH 2
2
2
2
11
1
1)(ωω
ωω
ωω
ω−−
−+
−=
CL
CL
ZRZjH+
=)( ω
At Very Low Frequencies
2)(
00)(
)(
πω
ωω
ωω
=∠
=→
=
jH
jHR
LjjH
LOW
LOW
LOW
At Very High Frequencies
2)(
01)(
)( 2
πω
ωω
ωωωω
−=∠
=∞
=∞→
−=
−=
jH
jH
RCj
LRCLjjH
HIGH
HIGH
HIGH
At the Resonant Frequency
11)11(
1
)( 20 =
+
−
=
LLC
jLCLC
R
LLC
jjH ω
0)(1)( 00 =∠= ωω jHjH
LC1
0 =ω
Our example is a band pass filter
Phase
φ = 90 at ω → 0
φ = 0 at ω0
φ = -90 at ω → ∞
Magnitude
Η = 0 at ω → 0
H=1 at ω0
Η = 0 at ω → ∞
f0